p39_081

# p39_081 - T T = 2 k L = 2(6 . 67 10 9 m 1 )(0 . 010)(750 10...

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81. (a) If m is the mass of the particle and E is its energy, then the transmission coeﬃcient for a barrier of height U and width L is given by T = e 2 kL , where k = r 8 π 2 m ( U E ) h 2 . If the change ∆ U in U is small (as it is), the change in the transmission coeﬃcient is given by T = dT dU U = 2 LT dk d U U. Now, dk dU = 1 2 U E r 8 π 2 m h 2 = 1 2( U E ) r 8 π 2 m ( U E ) h 2 = k 2( U E ) . Thus, T = LTk U U E . For the data of Sample Problem 39-7, 2 kL =10 . 0, so kL =5 . 0and T T = kL U U E = (5 . 0) (0 . 010)(6 . 8eV) 6 . 8eV 5 . 1eV = 0 . 20 . There is a 20% decrease in the transmission coeﬃcient. (b) The change in the transmission coeﬃcient is given by T = dT dL L = 2 ke 2 kL L = 2 kT L and
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Unformatted text preview: T T = 2 k L = 2(6 . 67 10 9 m 1 )(0 . 010)(750 10 12 m) = . 10 . There is a 10% decrease in the transmission coecient. (c) The change in the transmission coecient is given by T = dT dE E = 2 Le 2 kL dk dE E = 2 LT dk dE E . Now, dk/dE = dk/dU = k/ 2( U E ), so T T = kL E U E = (5 . 0) (0 . 010)(5 . 1 eV) 6 . 8 eV 5 . 1 eV = 0 . 15 . There is a 15% increase in the transmission coecient....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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